Problem in evaluating logarithm derivatives












8














Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question
























  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    yesterday










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    yesterday












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    yesterday










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
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8














Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question
























  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    yesterday










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    yesterday












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    yesterday










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    yesterday














8












8








8


1





Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms







share|cite|improve this question















Given the property of the logarithm that $log{xy} = log{x} + log{y}$, how would one take the 'derivative' of this?



To be more clear,



$log{xy} = log{x} + log{y}$ (property of $log$)



$D(log{xy}) = D(log{x} + log{y})$ (i) (Take derivative on both sides)



Now, $D(log{x} + log{y}) = D(log{x}) + D(log{y})$ (ii) (Derivative of sum is sum of derivatives)



Combining (i) and (ii): $D(log{xy}) = D(log{x}) + D(log{y})$ (iii)



Implies: $frac{1}{xy} = frac{1}{x} + frac{1}{y}$ (Evaluate derivative of logarithm using $D(log{x}) = frac{1}{x}$



$frac{1}{xy} = frac{1}{x} + frac{1}{y}$ looks false to me; e.g. while $log{6}$ does equal $log{2} + log{3}$, $frac{1}{6}$ does not equal $frac{1}{2} + frac{1}{3}$.



My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.



What's going wrong in this example?






real-analysis calculus logarithms




real-analysis calculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









David G. Stork

9,93521232




9,93521232










asked yesterday









user2192320

473




473












  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    yesterday










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    yesterday












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    yesterday










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    yesterday


















  • Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
    – MathIsLife12
    yesterday










  • (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
    – Clement C.
    yesterday












  • Thanks guys; this clears things up! I appreciate the fast responses.
    – user2192320
    yesterday










  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – John Doe
    yesterday
















Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
– MathIsLife12
yesterday




Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal.
– MathIsLife12
yesterday












(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
– Clement C.
yesterday






(This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.)
– Clement C.
yesterday














Thanks guys; this clears things up! I appreciate the fast responses.
– user2192320
yesterday




Thanks guys; this clears things up! I appreciate the fast responses.
– user2192320
yesterday












After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– John Doe
yesterday




After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– John Doe
yesterday










2 Answers
2






active

oldest

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21














Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
So there are no inconsistencies.






share|cite|improve this answer





















  • Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    yesterday










  • No problem, happy to help!
    – John Doe
    yesterday








  • 2




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    23 hours ago



















10














The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
$$
At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






share|cite|improve this answer








New contributor




ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    21














    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer





















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      yesterday










    • No problem, happy to help!
      – John Doe
      yesterday








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      23 hours ago
















    21














    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer





















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      yesterday










    • No problem, happy to help!
      – John Doe
      yesterday








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      23 hours ago














    21












    21








    21






    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.






    share|cite|improve this answer












    Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$frac{partial log(xy)}{partial x}=frac{y}{xy}=frac1x$$and $$frac{partial}{partial x}(log x+log y)=frac1x+0=frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).



    Alternatively, you could take a total derivative. $$mathrm d(log xy)=frac{partial log xy}{partial x}mathrm dx+frac{partial log xy}{partial y}mathrm dy=frac1xmathrm dx+frac1y mathrm dy$$This agrees with $$d(log x+log y)=frac1xmathrm dx+frac1y mathrm dy$$
    So there are no inconsistencies.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    John Doe

    10.4k11135




    10.4k11135












    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      yesterday










    • No problem, happy to help!
      – John Doe
      yesterday








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      23 hours ago


















    • Thank you for the quick response; this is my favorite answer posted so far.
      – user2192320
      yesterday










    • No problem, happy to help!
      – John Doe
      yesterday








    • 2




      @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
      – Ruslan
      23 hours ago
















    Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    yesterday




    Thank you for the quick response; this is my favorite answer posted so far.
    – user2192320
    yesterday












    No problem, happy to help!
    – John Doe
    yesterday






    No problem, happy to help!
    – John Doe
    yesterday






    2




    2




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    23 hours ago




    @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer.
    – Ruslan
    23 hours ago











    10














    The issue is that your differential operator $D$ does not behave in the way you think it does!
    $$
    D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
    $$
    At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






    share|cite|improve this answer








    New contributor




    ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      10














      The issue is that your differential operator $D$ does not behave in the way you think it does!
      $$
      D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
      $$
      At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






      share|cite|improve this answer








      New contributor




      ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        10












        10








        10






        The issue is that your differential operator $D$ does not behave in the way you think it does!
        $$
        D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
        $$
        At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.






        share|cite|improve this answer








        New contributor




        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        The issue is that your differential operator $D$ does not behave in the way you think it does!
        $$
        D(log(xy))=frac{1}{xy}D(xy)=frac{1}{xy}(ydx+xdy)=frac{1}{x}dx+frac{1}{y}dy=D(log(x))+D(log(y))
        $$
        At no point in this computation do we actually have $frac{1}{xy}=frac{1}{x}+frac{1}{y}$ -- we are working with differentials, and not derivatives.







        share|cite|improve this answer








        New contributor




        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered yesterday









        ItsJustTranscendenceBro

        1712




        1712




        New contributor




        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        New contributor





        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        ItsJustTranscendenceBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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            Display a custom attribute below product name in the front-end Magento 1.9.3.8

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            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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